In the following monobromination reaction,the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The starting material is $2$-bromobutane,which is enantiomerically pure. Monobromination can occur at different positions:$1.$ Substitution at $C-

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) The starting material is $2$-bromobutane,which is enantiomerically pure. Monobromination can occur at different positions:$1.$ Substitution at $C-2$: This forms $2,2$-dibromobutane,which is achiral.$2.$ Substitution at $C-1$: This forms $1,2$-dibromobutane. The $C-2$ is already chiral. The new chiral center at $C-1$ creates two diastereomers ($R,R$ and $S,R$ configurations). Both are chiral.$3.$ Substitution at $C-3$: This forms $2,3$-dibromobutane. The $C-2$ is already chiral. The new chiral center at $C-3$ creates two diastereomers. Both are chiral.$4.$ Substitution at $C-4$: This forms $1,4$-dibromobutane (not possible here as it is $2$-bromobutane). Wait,the structure is $2$-bromopentane. Let's re-evaluate: The reactant is $2$-bromopentane. Positions for substitution: $C-1$: $1,2$-dibromopentane (chiral,$2$ diastereomers).$C-2$: $2,2$-dibromopentane (achiral).$C-3$: $2,3$-dibromopentane (chiral,$2$ diastereomers).$C-4$: $2,4$-dibromopentane (chiral,$2$ diastereomers).$C-5$: $1,5$-dibromopentane (chiral,$2$ diastereomers).However,based on the provided image options and standard radical stability,the chiral products are formed at $C-1, C-3, C-4, C-5$. Each gives a pair of diastereomers. The question asks for the number of chiral products. Given the options,the correct answer is $4$.

In the following monobromination reaction,the number of possible chiral products is

Similar Questions

The reaction of lithium di-$3-$pentylcuprate with ethyl bromide yields an alkane with the molecular formula $C_7H_{16}$. What is the structure of the alkane?

$A$ mixture of all isomers possible from the mono-chlorination of the structure $n$-pentane $(CH_3-CH_2-CH_2-CH_2-CH_3)$ is subjected to fractional distillation. How many fractions will be obtained?

The major products $P$ and $Q$ of the following reactions respectively are:
$I$. $n$-pentyl bromide $\xrightarrow{Zn / H^{+}} P$
$II$. $n$-pentyl bromide $\xrightarrow[\text{Dry ether}]{Na} Q$

Why is direct iodination of alkanes not possible?

In which of the following reactions is the product a hydrocarbon?

Vedclass Test Series

Exam Paper Generator

Online Exam Module